Answer:
Proved
Step-by-step explanation:
To prove that every point in the open interval (0,1) is an interior point of S
This we can prove by contradiction method.
Let, if possible c be a point in the interval which is not an interior point.
Then c has a neighbourhood which contains atleast one point not in (0,1)
Let d be the point which is in neighbourhood of c but not in S(0,1)
Then the points between c and d would be either in (0,1) or not in (0,1)
If out of all points say d1,d2..... we find that dn is a point which is in (0,1) and dn+1 is not in (0,1) however large n is.
Then we find that dn is a boundary point of S
But since S is an open interval there is no boundary point hence we get a contradiction. Our assumption was wrong.
Every point of S=(0, 1) is an interior point of S.
1. No
2. No
3. Yes
4. No, it’s actually 24j -36
Answer:
D. d = sqrt( ( x2-x1)^2 + (y2-y1)^2)
C. d = sqrt( ( x1-x2)^2 + (y1-y2)^2)
B. d = sqrt( (|x2-x1|^2 + |y2-y1|^2)
Step-by-step explanation:
Given two points (x1, y1) and (x2,y2) we can find the distance using
d = sqrt( ( x2-x1)^2 + (y2-y1)^2)
The order of the terms inside the square doesn't matter
d = sqrt( ( x1-x2)^2 + (y1-y2)^2)
When we are squaring are term, we can take the absolute value before we square and it does not change the value
d = sqrt( (|x2-x1|^2 + |y2-y1|^2)
Answer:x
Step-by-step explanation:
Answer:
Its d 5/10
Step-by-step explanation:
